An image reading apparatus that is generally incorporated in a facsimile machine, a scanner, or the like is well known as a linear image reading apparatus, in which a transparent plate such as a glass plate is fixed by an adhesive over a row of photoelectric elements (refer to Japanese Patent Laid Open No. Sho. 63-9358 (1988), for instance).
FIG. 10 shows schematically the optical arrangement of a conventional linear image reading apparatus with the same basic configuration as above; FIG. 10 (a) is a schematic vertical section illustrating the boundary case of a ray of light originating from scattering at the surface of the photoelectric elements, and FIG. 10 (b) is a schematic plan view illustrating the extent of the portion of photoelectric elements 20 subjected to secondary incidence of light rays totally reflected at an upper boundary separating a protective plate and the outside air.
The conventional image reading apparatus shown in FIG. 10 includes a plurality of photoelectric elements 20 aligned in the transverse direction in the figure, and a glass plate 21 fixed by an adhesive layer 22 over the photoreceptive face of photoelectric elements 20 (the upper side in the figure). The glass plate 21 protects photoelectric elements 20 from both humidity and scanned originals so as not to be worn nor damaged by them, which consequently allows the image reading apparatus to maintain its resolution.
To describe an optical path in the image reading apparatus with the above configuration, an incident ray of light normal to glass plate 21 is taken for the sake of simplicity as an example. The incident ray of light normal to glass plate 21 (the ray indicated by a solid line in FIG. 10 (a)) goes straight through glass plate 21 and adhesive layer 22 of epoxy, for example, to impinge on the photoreceptive face 20a of photoelectric elements 20, where most of the ray is absorbed in photoelectric elements 20 and the remainder (indicated by a double-dotted line in the FIG. 10 (a)) is subject to diffuse back-scattering.
Here, for the purpose of sample calculations for tracing an optical path taken by the ray originating from scattering at the photoreceptive face 20a as described above, the thickness, d1, of adhesive layer 22 is taken as 15 .mu.m, the refractive indexes, n1 and n2, of adhesive layer 22 and glass plate 21 respectively as both 1.5, and the refractive index, n3, of air as 1. There is no reflection at boundary 9 separating adhesive layer 22 and glass plate 21, and the scattered ray continues in a straight line. The angle of incidence at which the ray in adhesive layer 22 strikes boundary 9 is therefore equal to the angle at which the ray is transmitted into glass plate 21, both of which are indicated by .theta..sub.1 in FIG. 10 (a). The scattered ray making the angle .theta..sub.1 to the normal goes straight in glass plate 21 to approach boundary 10 separating glass plate 21 and the outside air at the same angle. If this angle of incidence at boundary 10 is equal to or more than the critical angle, .theta. .sub.2, the ray is totally reflected at boundary 10 to return toward the photoelectric elements 20. The figure shows the boundary case of a ray of light scattered at the critical angle, .theta..sub.2, and returning to the photoreceptive face 20a at a point R2. This point R2 therefore delineates the inner boundary of the area exposed to secondarily incident rays. This boundary is indicated by a circle C shown in FIG. 10 (b), which is drawn using a point R1 as a center with a radius of Ld.sub.1, and the outside of circle C (the outside indicated by hatching in the figure) may receive light rays sent by total reflection. Here, R1 is assumed to be the point where the ray is first incident on photoreceptive face 20a and also where the scattered ray originated. To summarize, scattered rays of light originating at point R1 undergo total reflection at boundary 10 to return to positions outside circle C drawn round center R1 with radius Ld.sub.1.
Making the assumptions about the refractive indexes given above, the equation for calculating the critical angle, .theta..sub.2, for boundary 10 from which total reflection takes place, is expressed by:
1/sin.theta..sub.2 =n2/n3 (1)
where n2 is the refractive index of glass plate 21 and n3 is the refractive index of air.
Using the values given above, solving equation (1) for .theta..sub.2 gives .theta..sub.2 -41.8.degree. for the critical angle.
The isosceles triangle R1R2R3 shown in FIG. 10 (a) has as one side part of the transverse line representing the photoelectric elements 20 and as the other two, part of the double-dotted line representing the optical path taken by the boundary case of a scattered ray. From this triangle, elementary geometry yields the equation for calculating the distance Ld.sub.1 : EQU Ld.sub.1 =2(d1+d2)tan.theta..sub.2 ( 2)
where d1 is the thickness of adhesive layer and d2 is the thicknesses of glass plate 21.
Substitution of the values given above into equation (2) gives Ld.sub.1 =116.2 .mu.m for the distance.
As can be seen in FIG. 10 (b), two lines parallel with each other extend transversely, distance L1 apart, to form a strip-shaped area where photoelectric elements 20 are aligned end to end in the transverse direction (i.e. the fast scan direction). Accordingly, the distance L1 corresponds to the dimension of a side of a photoelectric element 20 in the slow scan direction orthogonal to the alignment direction. As shown in the figure, the strip-shaped area of width L1 extends into the hatched area, which means that some light rays subject to total reflection at boundary 10 are included there. The amount of scattered light falling within the strip-shaped area of width L1 can be represented as a proportion to the total light scattered at the critical angle or more. For the purpose of sample calculations of the amount of light, L1=63.5 .mu.m and Ld.sub.1 =116.2 .mu.m are assumed. The value of 63.5 .mu.m is equivalent to the dimension of a side of a photoelectric element 20 corresponding to a picture element density of 400 dots per inch. The strip-shaped area of width L1 intersects the circumference of circle C to form two circular arcs 23 and 24, each of which subtends a central angle .phi.1. Since L1=2.times.Ld1.times.sin(.phi.1/2), substitution gives .phi.1.times.31.7.degree. for the central angle. As mentioned above, the amount of light included in the area of width L1 can be represented as the proportion A of the total light scattered at the critical angle or more. The proportion A is roughly approximate to the ratio of twice the center angles .phi.1 to all directions, so that A=2.phi.1/360=0.18.
Thus, eighteen percent of the totally reflected rays at boundary 10 separating glass plate 21 and outside air, are incident again on photoelectric elements 20, to interfere with direct incident rays from the outside onto photoelectric elements 20, consequently resulting in lower resolving power of the image reading apparatus and image quality of the output image.